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TOPICS
Exercise - 5.2

Question-1 :-  Fill in the blanks in the following table, given that a is the first term, d the common difference and an the nth term of the AP:

Solution :-
```(i) Given : a = 7, d = 3, n = 8, an = ?
an = a + (n-1)d
a8 = 7 + (8-1)3
a8 = 7 + 21
a8 = 28
```
```(ii) Given : a = -18, d = ?, n = 10, an = 0
an = a + (n-1)d
0 = -18 + (10-1)d
18 = 9d
d = 18/9
d = 2
```
```(iii) Given : a = ?, d = -3, n = 18, an = -5
an = a + (n-1)d
-5 = a + (18-1)(-3)
-5 = a - 51
a = -5 + 51
a = 46
a8 = 28
```
```(iv) Given : a = -18.9, d = 2.5, n = ?, an = 3.6
an = a + (n-1)d
3.6 = -18.9 + (n-1)(2.5)
3.6 = -18.9 + 2.5n - 2.5
2.5n = 3.6 + 18.9 + 2.5
2.5n = 25.0
n = 25/2.5
n = 10
```
```(v) Given : a = 3.5, d = 0, n = 105, an = ?
an = a + (n-1)d
a105 = 3.5 + (105-1)0
a105 = 3.5 + 0
a105 = 3.5
```

Question-2 :-  Choose the correct choice in the following and justify :
(i) 30th term of the AP: 10, 7, 4, . . . , is
(A)97   (B) 77   (C) –77   (D) – 87

(ii) 11th term of the AP: – 3, -1/2 , 2, . . ., is
(A) 28   (B) 22   (C) –38   (D) – 48 by 1/2

Solution :-
```(i) Here, a = 10, d = 7 - 10 = -3, n = 30
an = a + (n-1)d
a30 = 10 + (30-1)(-3)
a30 = 10 - 87
a30 = -77
Hence, the correct answer is C.
```
```(ii) Here, a = -3, d = -1/2 + 3 = 5/2, n = 11
an = a + (n-1)d
a11 = -3 + (11-1)(5/2)
a11 = -3 + 25
a11 = 22
Hence, the correct answer is B.
```

Question-3 :-  In the following APs, find the missing terms in the boxes :
(i) 2, ___, 26
(ii) ___, 13, ___, 3
(iii) 5, ___, ___, 9 by 1/2
(iv) -4, ___, ___, ___, ___, 6
(v) ___, 38, ___, ___, ___, -22

Solution :-
```(i) For this A.P., a = 2, a3 = 26
We know that, an = a + (n−1)d
a3 = 2 + (3−1)d
26 = 2 + 2d
24 = 2d
d = 12
a2 = 2 + (2−1)12 = 14
Therefore, 14 is the missing term.
```
```(ii) For this A.P.,
a2 = 13 and a4 = 3
We know that, an = a + (n−1)d
a2 = a + (2−1)d
13 = a + d ..... (1)
a4 = a + (4−1)d
3 = a + 3d .... (2)
By subtracting (1) from (2), we obtain
−10 = 2d
d = −5
From equation (1), we obtain
13 = a + (−5)
a = 18
a3 = 18 + (3−1)(−5) = 18 + 2 x (−5) = 18 − 10 = 8
Therefore, the missing terms are 18 and 8 respectively.
```
```(iii) For this A.P.,
a = 5 and a4 = 9 by 1/2 = 19/2
We know that
an = a + (n−1)d
a4 = 5 + (4-1)d
19/2 = 5 + 3d
3d = 19/2 - 5
3d = 9/2
d = 9/6
d = 3/2
Now,
a2 = a + d = 5 + 3/2 = 13/2
a3 = a + 2d = 5 + 2 x 3/2 = 5 + 3 = 8
Therefore, the missing terms are 13/2 and 8 respectively.
```
```(iv) For this A.P.,
a = −4 and a6 = 6
We know that,
an = a + (n−1)d
a6 = a + (6−1)d
6 = − 4 + 5d
10 = 5d
d = 2
Now,
a2 = a + d = − 4 + 2 = −2
a3 = a + 2d = − 4 + 2 x 2 = 0
a4 = a + 3d = − 4 + 3 x 2 = 2
a5 = a + 4d = − 4 + 4 x 2 = 4
Therefore, the missing terms are −2, 0, 2, and 4 respectively.
```
```(v) For this A.P.,
a2 = 38 and a6 = −22
We know that
an = a + (n−1)d
a2 = a + (2−1)d
38 = a + d ....(1)
a6 = a + (6−1)d
−22 = a + 5d ....(2)
On subtracting equation (1) from (2), we obtain
−22 − 38 = 4d
−60 = 4d
d = −15
Now,
a = a2 − d = 38 − (−15) = 53
a3 = a + 2d = 53 + 2 x (−15) = 23
a4 = a + 3d = 53 + 3 x (−15) = 8
a5 = a + 4d = 53 + 4 x (−15) = −7
Therefore, the missing terms are 53, 23, 8, and −7 respectively.
```

Question-4 :-  Which term of the AP : 3, 8, 13, 18, . . . ,is 78?

Solution :-
```    Here, a = 3, d = 8 - 3 = 5, an = 78
an = a + (n-1)d
78 = 3 + (n-1)5
78 = 3 + 5n - 5
78 = 5n - 2
5n = 78 + 2
5n = 80
n = 80/5
n = 16
Therefore, 16th term is 78.
```

Question-5 :-  Find the number of terms in each of the following APs :
(i) 7, 13, 19, .... , 205
(ii) 18, 15 by 1/2, 13, .... , -47?

Solution :-
```(i) Here, a = 7, d = 13 - 7 = 6, an = 205
an = a + (n-1)d
205 = 7 + (n-1)6
205 = 7 + 6n - 6
205 = 6n + 1
6n = 205 - 1
n = 204/6
n = 34
```
```(ii) Here, a = 18, d = 15 by 1/2 - 18 = 31/2 - 18 = -5/2, an = -47
an = a + (n-1)d
-47 = 18 + (n-1)(-5/2)
-47 = 18 - 5n/2 + 5/2
5n/2 = 41/2 + 47
5n/2 = 135/2
n = 135/5
n = 27
```

Question-6 :-  Check whether – 150 is a term of the AP : 11, 8, 5, 2 ....

Solution :-
```    Here, a = 11, d = 8 - 11 = -3, an = -150
an = a + (n-1)d
-150 = 11 + (n-1)(-3)
-150 = 11 - 3n + 3
3n = 14 + 150
3n = 164
n = 164/3
Therefore, n is not an integer.
So, -150 is not a term of this A.P.
```

Question-7 :-  Find the 31st term of an AP whose 11th term is 38 and the 16th term is 73.

Solution :-
```    According to question, we have
a11 = a + 10d = 38 ....(1)
a16 = a + 15d = 73 ....(2)
By solving, (1) & (2) equation
a + 10d - a - 15d = 38 - 73
-5d = -35
d = 35/5
d = 7
Put the value of d = 7 in (1)equation, we find
a + 10 x 7 = 38
a + 70 = 38
a = 38 - 70
a = -32

Now, a31 = ?
a31 = a + 30d = -32 + 30 x 7 = -32 + 210 = 178
Therefore, 31st term is 178.
```

Question-8 :-  An AP consists of 50 terms of which 3rd term is 12 and the last term is 106. Find the 29th term.

Solution :-
```    According to question, we have
a3 = a + 2d = 12 ....(1)
an = a + (n-1)d = 106 ....(2)
Here, A.P. consits of 50 terms. So, n = 50
Put n = 50 in (2) equation, we find
a50 = a + (50-1)d = a + 49d = 106 ....(3)
By solving, (1) & (3) equation
a + 2d - a - 49d = 12 - 106
-47d = -94
d = 94/47
d = 2
Put the value of d = 2 in (1), we find
a + 2 x 2 = 12
a + 4 = 12
a = 12 - 4
a = 4

Now, a29 = ?
a29 = a + 28d = 4 + 28 x 2 = 4 + 56 = 60
```

Question-9 :-  If the 3rd and the 9th terms of an AP are 4 and – 8 respectively, which term of this AP is zero?

Solution :-
```    According to question, we have
a3 = a + 2d = 4 ....(1)
a9 = a + 8d = -8 ....(2)
By solving, (1) & (2) equation
a + 2d - a - 8d = 4 + 8
-6d = 12
d = -12/6
d = -2
Put the value of d = -2 in (1)equation, we find
a + 2 x (-2) = 4
a - 4 = 4
a = 4 + 4
a = 8

Now, an = 0
a + (n-1)d = 0
8 + (n-1)(-2) = 0
8 - 2n + 2 = 0
10 - 2n = 0
-2n = -10
n = 10/2
n = 5
Therefore, 5th term is Zero.
```

Question-10 :-  The 17th term of an AP exceeds its 10th term by 7. Find the common difference.

Solution :-
```    According to question, we have
a17 = a10 + 7
a + 16d = a + 9d + 7
16d - 9d = 7
7d = 7
d = 7/7
d = 1
Therefore, the common difference is 1.
```

Question-11 :-  Which term of the AP : 3, 15, 27, 39, . . . will be 132 more than its 54th term?

Solution :-
```    According to question, we have
3, 15, 27, 39, . . . = a54 + 132
a = 3, d = 15 - 3 = 12
an = a + (n-1)d
an = 3 + (n-1)12
an = 3 + 12n - 12
an = 12n - 9
12n - 9 = a54 + 132
12n - 9 = a + 53d + 132
12n - 9 = 3 + 53 x 12 + 132
12n = 135 + 636 + 9
12n = 780
n = 780/12
n = 65
Therefore, 65th term is 132 more than its 54th term.
```

Question-12 :-  Two APs have the same common difference. The difference between their 100th terms is 100, what is the difference between their 1000th terms?

Solution :-
```    Let the first term of these A.P.s be a1 and a2 respectively and the common difference of these A.P.s be d.
For first A.P.,
a100 = a1 + (100 − 1) d = a1 + 99d
a1000 = a1 + (1000 − 1) d = a1 + 999d

For second A.P.,
a100 = a2 + (100 − 1) d = a2 + 99d
a1000 = a2 + (1000 − 1) d = a2 + 999d
Given that, difference between 100th term of these A.P.s = 100
Therefore, (a1 + 99d) − (a2 + 99d) = 100

a1 − a2 = 100 ....(1)
Difference between 1000th terms of these A.P.s

(a1 + 999d) − (a2 + 999d) = a1 − a2
From equation (1),
This difference, a1 − a2 = 100
Hence, the difference between 1000th terms of these A.P. will be 100.
```

Question-13 :-  How many three-digit numbers are divisible by 7?

Solution :-
```    According to question, Divisible of 7 three-digit numbers are 105, 112, 119 ......
The maximum possible three-digit number is 999. When we divide it by 7, the remainder will be 5.
Clearly, 999 − 5 = 994 is the maximum possible three-digit number that is divisible by 7.

105, 112, 119 ...... 994
a  = 105, d = 112 - 105 = 7, an = 994
an = a + (n-1)d
994 = 105 + (n-1)7
994 = 105 + 7n - 7
994 = 98 + 7n
7n = 994 - 98
7n = 896
n = 896/7
n = 128
Therefore, 128 three-digit numbers are divisible by 7.
```

Question-14 :-  How many multiples of 4 lie between 10 and 250?

Solution :-
```    According to question, multiples of 4 lie between 10 and 250 are 12, 16, 20, 24, ....
When we divide 250 by 4, the remainder will be 2.
Therefore, 250 − 2 = 248 is divisible by 4.

12, 16, 20, 24 ...... 248
a = 12, d = 16 - 12 = 4, an = 248
an = a + (n-1)d
248 = 12 + (n-1)4
248 = 12 + 4n - 4
248 = 8 + 4n
4n = 248 - 8
4n = 240
n = 240/4
n = 60
Therefore, there are 60 multiples of 4 between 10 and 250.
```

Question-15 :-  For what value of n, are the nth terms of two APs: 63, 65, 67, . . . and 3, 10, 17, . . . equal?

Solution :-
```    Firstly, A.P. : 63, 65, 67, ....
a = 63, d = 65 - 63 = 2
an = a + (n-1)d
an = 63 + (n-1)2
an = 63 + 2n - 2
an = 61 + 2n

Secondly, A.P. 3, 10, 17, ....
a = 3, d = 10 - 3 = 7
an = a + (n-1)d
an = 3 + (n-1)7
an = 3 + 7n - 7
an = 7n - 4

According to question, 63, 65, 67, . . . and 3, 10, 17, . . . equal
61 + 2n = 7n - 4
2n - 7n = -4 - 61
-5n = -65
n = 65/5
n = 13
Therefore, 13th terms of both these A.P.s are equal to each other.
```

Question-16 :-  Determine the AP whose third term is 16 and the 7th term exceeds the 5th term by 12.

Solution :-
```    According to question, we have
a3 = a + 2d = 16 ....(1)
a7 = a5 + 12
a + 6d = a + 4d + 12
6d - 4d = 12
2d = 12
d = 12/2
d = 6
Put the value of d = 6 in (1) equation
a + 2 x 6 = 16
a + 12 = 16
a = 16 - 12
a = 4
Therefore, A.P. will be 4, 10, 16, .....
```

Question-17 :-  Find the 20th term from the last term of the AP : 3, 8, 13, . . ., 253.

Solution :-
```    Here, d = 5
Therefore, this A.P. can be written in reverse order as 253, 248, 243, ...., 13, 8, 5
a = 253, d = 5, n = 20
an = a + (n-1)d
a20 = 253 + (20-1)5
a20 = 253 + 95
a20 = 348
Therefore, 20th term from the last term is 158.
```

Question-18 :-  The sum of the 4th and 8th terms of an AP is 24 and the sum of the 6th and 10th terms is 44. Find the first three terms of the AP.

Solution :-
```    According to question, we have
a4 + a8 = 24
a + 3d + a + 7d = 24
2a + 10d = 24
a + 5d = 12 ....(1)
a6 + a10 = 44
a + 5d + a + 9d = 44
2a + 14d = 44
a + 7d = 22 ....(2)
By solving, (1) & (2) equation
a + 5d - a - 7d = 24 - 22
2d = 10
d = 10/2
d = 5
Put the value of d = 1 in (1) equation
a + 5 x 5 = 12
a + 25 = 12
a = 12 - 25
a = -13
Therefore, the first three terms of A.P. are a1, a2, a3
a1 = a + 0d = -13
a2 = a + d = -13 + 5 = -8
a3 = a + 2d = -13 + 2 x 5 = -13 + 10 = -3
```

Question-19 :-  Subba Rao started work in 1995 at an annual salary of ₹5000 and received an increment of ₹200 each year. In which year did his income reach ₹7000?

Solution :-
```    It can be observed that the incomes that Subba Rao obtained in various years are in A.P. as every year,
his salary is increased by ₹200.
Therefore, the salaries of each year after 1995 are 5000, 5200, 5400, ....
Here, a = 5000, d = 200
Let after nth year, his salary be ₹7000.
Therefore, an = a + (n−1)d
7000 = 5000 + (n−1)200
200(n−1) = 2000
n − 1 = 10
n = 10 + 1
n = 11
Therefore, in 11th year, his salary will be ₹7000.
```

Question-20 :-  Ramkali saved ₹5 in the first week of a year and then increased her weekly savings by ₹1.75. If in the nth week, her weekly savings become ₹20.75, find n.

Solution :-
```    According to question, we have
a = 5, d = 1.75, an = 20.75, n = ?
an = a + (n-1)d
20.75 = 5 + (n-1)(1.75)
20.75 = 5 + 1.75n - 1.75
20.75 = 3.25 + 1.75n
1.75n = 20.75 - 3.25
1.75n = 17.50
n = 17.50/1.75
n = 10
Therefore, n = 10 week.
```
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